Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

class Dijkstra {
public:
    struct edge { long long v, dist; };

    struct state {
        long long v, cost;
        bool operator>(const state s) const { return cost > s.cost; }
    };

    const long long INF = (1LL << 60);
    long long N;
    vector< vector<edge> > E;

    Dijkstra(long long n): N(n), E(n) {}

    //有効グラフの時はこっち。u→vに距離dで結ぶ
    void add_directed_edge(long long u, long long v, long long d) {
        E[u].push_back((edge) { v, d });
    }

    //無向グラフの時はこっち。uとvを双方向に距離dで結ぶ
    void add_undirected_edge(long long u, long long v, long long d) {
        E[u].push_back((edge) { v, d });
        E[v].push_back((edge) { u, d });
    }

    //Sを始点として、他の頂点への最短経路を探す
    vector<long long> shortest_path(long long S) {
        vector<long long> dp(E.size(), INF);
        priority_queue<state, vector<state>, greater<state> > q;
        q.push((state) { S, 0 });
        while(!q.empty()) {
            long long v = q.top().v, cost = q.top().cost;
            q.pop();
            if(dp[v] <= cost) continue;
            dp[v] = cost;
            for(int i=0;i < E[v].size() ; i++) {
            long long nv = E[v][i].v, ncost = cost + E[v][i].dist;
            if(dp[nv] > ncost) q.push((state) { nv, ncost });
            }
        }
        return dp;
    }
};


int main(){
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    s--,t--;
    Dijkstra d(n);
    long long x,y;
    for(int i=0;i<m;i++){
        cin>>x>>y;
        x--,y--;
        d.add_undirected_edge(x,y,1);
    }
    vector<long long> vs=d.shortest_path(s);
    vector<long long> vt=d.shortest_path(t);
    long long dist=vs[t];
    map<long long,long long> cs,ct;
    for(int i=0;i<n;i++){
        cs[vs[i]]++;
        ct[vt[i]]++;
        //cerr<<"a "<<vs[i]<<" "<<vt[i]<<endl;
    }
    long long ans=0;
    for(int i=0;i<=n;i++){
        ans+=cs[i]*ct[dist-2-i];
        //cerr<<"b"<<" "<<cs[i]<<" "<<ct[dist-2-i]<<endl;
    }
    cout<<ans<<endl;

    

}